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Offline laddoo12

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Calorimeter constant
« on: March 17, 2012, 01:49:24 AM »
hey guys so I've done a lab and so far this is what I've done for my calculations


for energy required to melt ice you would take
The energy of fusion of ice given (6010J/mol x grams of ice/molormass of water), which for me would be 6010j/molx(8.56gofice/18) = 2858J

then I need the energy absorbed by "ice" water as it warms to tempreture final
so we know the heat of fusion being 6010J/mol ice and the tempreture final is 7.88 degrees and the grams of ice is 8.56g which i don't think i need,
and the specific heat of water is 4.184 J/gC  

so what i was thinking looking at the units we would do  
4.184J/gC x 7.88 C x 8.56 giving us the energy absorbed by the water as the ice cools? am i correct? which would be 282.22J

thus for energy lost by warm water as it cools it would just be Q=McDeltaT where we take 49.21gH20(4.184J/gC)(-13.20(tempreture lost by water)) and all the units cancel giving us the enrgy lost? which is =-2,718J

Then after this point i get lost at when your determining the energy lost by the calorimeter :S
it says a hint(what should all these heats add to give) wouldnt it just be the initial 0 then? and then energy lost by this colorimeter would just be 0?
because energy lost be ice is energy gained by water?

 
And the LAST QUESTION that i need help is, is with determining a calorimeter constant,

would i go with energy lost by water-enery of ice as water cools/the change in tepreture which would be

-2718-282/(-13.2 degree C) giving me 227J/C is this to BIG?!?!?! did i do something wrong with my calculations up top?!

Offline laddoo12

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Re: Calorimeter constant
« Reply #1 on: March 17, 2012, 03:02:18 AM »
well i found the energy lost which would be 2718 + calorimeter loss = 282,22 + 2858 (The heat of fusion was gained by the ice)
Calorimeter loss = 422.22 j

Would i just take that and devid it by 13.20 giving me C in J/C

or would i have to plug it into Q=mCdeltaT ??? cause if i did that putting the mass of a colorimeter i wouldn't have it in J/Celsuis ?

Offline Borek

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Re: Calorimeter constant
« Reply #2 on: March 17, 2012, 06:17:27 AM »
Help us help you. Please post procedure you followed and data you collected, then explain what is the problem. So far all I see is a chaotic mix of numbers and calculations, forcing me to guessing what is what of what.
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Offline laddoo12

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Re: Calorimeter constant
« Reply #3 on: March 17, 2012, 01:05:16 PM »
DATA
mass of calorimeter = 3.29
initial temperature = 21.08C
final tempreture = 7.88
mass of ice = 8.56
mass of water = 49.21
energy of fusion ice = 6010J/mole
C of water = 4.184 J/gC
need to find the colorimeter constant C in Joules/Celsius

Offline laddoo12

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Re: Calorimeter constant
« Reply #4 on: March 17, 2012, 01:06:34 PM »
the procedure is you put an ice cube into 50grams of water and measure the initial temperature and then the final temperature and you need to determine the calorimeter constant from those results!

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